Question

Given below are the steps involved in the generation and conduction of nerve impulse. Arrange them in proper sequence and select the correct option.
$I.$ Rapid influx of $N{a}^{+}$ ions followed by reversal of polarity at the site of application of stimulus.
$II.$ Reversal of polarity and generation of action potential at a point ahead of the point of application of stimulus.
$III.$ Development of free permeability to $N{a}^{+}$ ions at the site of application of stimulus.
$IV.$ Arrival of the impulse generated at the first site to the second site.
$V.$ Rise in permeability of $K^{+}$ ions and restoration of the resting potential of the membrane.

• A. $V,II,III,IV,I$
• B. $I,III,II,V,IV$
• C. $III,II,I,IV,V$
• D. $III,I,II,IV,V$

Answer 1

Answer: (D)

The sequential order of the generation and conduction of nerve impulse is as follows:
$1.$ As soon as stimulus is applied at a site, (site A) the site becomes freely permeable to $N{a}^{+}$ ions.
$2.$ This causes a rapid influx of $N{a}^{+}$ ions intracellularly. This causes a reversal of polarity i.e. the intracellular surface acquires a positive charge and the extracellular surface acquires a negative charge.
$3.$ At a site ahead of the first site (site B), the extracellular surface has a positive and the intracellular surface has a negative charge. Thus, extracellularly, the current flows from site B to site A and intracellularly, it flows from site $A$ to site $B$ to complete the circuit. This causes a reversal of polarity at site $B$ .
$4.$ The impulse generated at site A is transmitted to site $B$ .
$5.$ The stimulus induced permeability to $N{a}^{+}$ ions disappears very quickly. The $K^{+}$ ions are diffused outside the cell and the resting potential of the membrane is restored.